Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction ratios of the given line are $1,2,2$.

The direction ratios of the normal to the given plane are $2,-1,\sqrt{\lambda }$.

The angle $\theta$ between a line and a plane is given by:

$\sin \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Substituting the respective values:

$\sin \theta =\frac{|(1)(2)+(2)(-1)+(2)(\sqrt{\lambda })|}{\sqrt{1^2+2^2+2^2}\sqrt{2^2+(-1{)}^2+(\sqrt{\lambda }{)}^2}}$

$\sin \theta =\frac{|2-2+2\sqrt{\lambda }|}{\sqrt{9}\sqrt{5+\lambda }}$

$\sin \theta =\frac{2\sqrt{\lambda }}{3\sqrt{5+\lambda }}$

Given that $\sin \theta =\frac{1}{3}$, we have:

$\frac{2\sqrt{\lambda }}{3\sqrt{5+\lambda }}=\frac{1}{3}$

$2\sqrt{\lambda }=\sqrt{5+\lambda }$

Squaring both sides:

$4\lambda =5+\lambda$

$3\lambda =5\Rightarrow \lambda =\frac{5}{3}$

Therefore, the value of $\lambda +1$ is:

$\lambda +1=\frac{5}{3}+1=\frac{8}{3}$

Answer: $\frac{8}{3}$